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October 21st, 2013, 07:43 AM   #1
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Approach reals by 3-smooth numbers

Hello everyone,

I am not a mathematician but I am looking for a theorem. I want to know if it is possible to approximate any real number by the quotient of two 3-smooth numbers. In practice that means: can any real number be approached arbitrary close by where m and n are in ?

Thanks in advance!!

Maurice
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October 21st, 2013, 08:03 AM   #2
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Re: Approach reals by 3-smooth numbers

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Originally Posted by Maurice1969
I want to know if it is possible to approximate any real number by the quotient of two 3-smooth numbers.
Well, I don't know of any such theorem, but I doubt whether rationals of the form {1/(2^m * 3^n), 2^m/3^n, 2^m * 3^n} for m,n in N is dense enough in R to make that true.
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October 21st, 2013, 10:15 AM   #3
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Re: Approach reals by 3-smooth numbers

Thanks for your reply. I doubt it also that every real number can be approached that way, but I really like to know how dense the set is and what kind of set is created by and m,n , maybe a kind of Cantor set. All info is welcome.

Maurice
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October 21st, 2013, 12:57 PM   #4
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Re: Approach reals by 3-smooth numbers

Well, this isn't a proof, but I believe you can approximate any real number arbitrarily closely with because you can get arbitrarily close to 1. is irrational, and hence you can approximate it arbitrarily closely with rational numbers, which translate into the values of m and n. The closer the ratio m:n is to , the closer and are to 1. Remember, you can approximate arbitrarily closely with rational numbers. Now suppose you want to approximate a real number x which is larger than 1. Take an approximation which is sufficiently close to 1 and greater than 1, and then continually multiply it by itself until you get a number which is larger than x, and call the exponent p. Then will be sufficiently close to x. If the real number x you wish to approximate is less than 1, then simply choose an approximation which is sufficiently close to 1 and less than 1, and continually multiply it by itself until you get a number which is less than x. In particular, the margin of error (percentage wise) in this calculation will be less than or equal to , which can be arbitrarily close to 0.
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October 21st, 2013, 01:16 PM   #5
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Re: Approach reals by 3-smooth numbers

Well obviously the numbers of the form are all positive and can thus only be dense in the nonnegative real numbers, but that is true. And in any case the full statement on 3-smooth numbers is true.

Unfortunately, i don't know of any bigger theorem implying it, so here's the outline of the proof: (It is rather sketchy, but if required I can fill the gaps later)

Step 1: It is enough to show: for any there are integral m, n such that .

In fact, assuming the assertion: For every real r > 1 and every take f = 2^m 3^m to be a number as described in the assertion, i.e. 1 < f < 1 + e. Then there is an integer k such that . Letting epsilon go to 0 we can thus approximate r. The proof for 0< r < 1 is similar.

Step 2: It is enough to show: for any there are integral m, n such that .

To prove the assertion of step 1 for any given assume the assertion of step 2 is valid . Then the same pair (m,n) will satisfy the conditions of step 1.

Step 3: The assertion of step 2 is well known to be true for any two nonzero reals a,b in stead of log( 2) and log (3) such that the quotient a/b is irrational. There is a nice proof using topological group theory, but i don't know a reference at the moment. maybe someone else knows one.
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October 21st, 2013, 04:04 PM   #6
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Re: Approach reals by 3-smooth numbers

Hi,
Yes any number can be approached in this way. As Peter as stated this is equivalent to ask if is dense in R. Note that this space forms a group, which is a subgroup of R. It is easy to show that any subgroups of R are either of the form with a is a real number or are dense in R. Obviously there is no such a in our case, the group has no smallest elements therefor this group is dense in R and any real number can be approached.
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October 21st, 2013, 11:12 PM   #7
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Re: Approach reals by 3-smooth numbers

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Step 2: It is enough to show: for any there are integral m, n such that .
.
Yes, I thought of that after I have questioned it's density, that the density of n log 2 + m log 3 might suffice, but haven't had the time to get through a formal outline.

So, definitely, nice thinking and nice proof, Peter.
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October 22nd, 2013, 06:00 AM   #8
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Re: Approach reals by 3-smooth numbers

Thank you all for the replies! It helped me a lot. Is it possible to say now that every positive real can be approached by with m and n in since it can be very close to one.

Thanks again!
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October 22nd, 2013, 07:31 AM   #9
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Re: Approach reals by 3-smooth numbers

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Is it possible to say now that every positive real can be approached by with m and n in since it can be very close to one.
That'd be false. For example, take any real in the interval [0, 1].
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October 22nd, 2013, 01:34 PM   #10
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Re: Approach reals by 3-smooth numbers

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Originally Posted by Maurice1969
Thank you all for the replies! It helped me a lot. Is it possible to say now that every positive real can be approached by with m and n in since it can be very close to one.

Thanks again!
My understanding is that this statement is true, but the reason it is true is thus: can be arbitrarily close to 1 (very close is not good enough) because is irrational (which is always true for two numbers which are relatively prime). Being arbitrarily close means you can be within any positive distance from 1 on the number line. Note that this statement is also true for the fraction , where m and n are natural numbers. It is not true for where m and n are natural numbers, since the only numbers obtainable in this fashion are integers, which cannot be arbitrarily close to 1.
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